From ebd7dd5bfbada4a66f18e27a18ae1fddb2daf0bc Mon Sep 17 00:00:00 2001
From: Jaron Kent-Dobias <jaron@kent-dobias.com>
Date: Mon, 7 Dec 2020 14:15:51 +0100
Subject: Added lazy paragraph defending our choice of constraint by smearing
 the alternative.

---
 bezout.tex | 8 +++++++-
 1 file changed, 7 insertions(+), 1 deletion(-)

(limited to 'bezout.tex')

diff --git a/bezout.tex b/bezout.tex
index 02158a0..897e79d 100644
--- a/bezout.tex
+++ b/bezout.tex
@@ -31,7 +31,7 @@
 \begin{equation} \label{eq:bare.hamiltonian}
   H_0 = \frac1{p!}\sum_{i_1\cdots i_p}^NJ_{i_1\cdots i_p}z_{i_1}\cdots z_{i_p},
 \end{equation}
-where the $z$ are constrained by $z\cdot z=N$ and $J$ is a symmetric tensor
+where $z\in\mathbb C^N$ is constrained by $z^2=N$ and $J$ is a symmetric tensor
 whose elements are complex normal with $\langle|J|^2\rangle=p!/2N^{p-1}$ and
 $\langle J^2\rangle=\kappa\langle|J|^2\rangle$ for complex parameter
 $|\kappa|<1$. The constraint is enforced using the method of Lagrange
@@ -41,6 +41,12 @@ multipliers: introducing the $\epsilon\in\mathbb C$, this gives
 \end{equation}
 At any critical point $\epsilon=H/N$, the average energy.
 
+When compared with $z^*z=N$, the constraint $z^2=N$ may seem an unnatural
+extension of the real $p$-spin spherical model. However, a model with this
+nonholomorphic spherical constraint has a disturbing lack of critical points
+nearly everywhere, since $0=\partial H/\partial z^*=-p\epsilon z$ is only
+satisfied for $\epsilon=0$, as $z=0$ is forbidden by the constraint.
+
 Since $H$ is holomorphic, a point is a critical point of its real part if and
 only if it is also a critical point of its imaginary part. The number of
 critical points of $H$ is therefore the number of critical points of
-- 
cgit v1.2.3-70-g09d2

